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M12-13

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M12-13  [#permalink]

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New post 16 Sep 2014, 00:46
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A
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C
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Re M12-13  [#permalink]

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New post 16 Sep 2014, 00:46
Official Solution:


Notice that we can get the equation of line \(k\) which is perpendicular to line \(y=2x\) if we know ANY point that line \(k\) passes through. So, to get the equation of line \(k\) we need the values of \(a\) and \(b\).

(1) \(a = -b\). Not sufficient.

(2) \(a - b = 1\). Not sufficient.

(1)+(2) \(a = -b\) and \(a - b = 1\), we have two distinct linear equations with two unknowns so we can solve for \(a\) and \(b\). Sufficient.


Answer: C
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Re: M12-13  [#permalink]

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New post 24 Nov 2015, 01:53
hi Bunuel

can't A suffice to find solution? y=2x is given as going from quad 3 through origin to quad 1 with slope 2.

slope of the perped line is -1/2, relatioship between points is given as whenever a is positive the b bears the same value but with opposite sign (10,-10 or -10;10) - hence the second line has to go through the origin from quad 2 to quad 4?

thanks
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Re: M12-13  [#permalink]

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New post 24 Nov 2015, 10:24
shasadou wrote:
hi Bunuel

can't A suffice to find solution? y=2x is given as going from quad 3 through origin to quad 1 with slope 2.

slope of the perped line is -1/2, relatioship between points is given as whenever a is positive the b bears the same value but with opposite sign (10,-10 or -10;10) - hence the second line has to go through the origin from quad 2 to quad 4?

thanks


Why should line k go through the origin?
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Re: M12-13  [#permalink]

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New post 02 Jul 2016, 03:38
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)
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Re: M12-13  [#permalink]

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New post 02 Jul 2016, 05:17
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)


Check the diagram below:

Image

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.


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Re M12-13  [#permalink]

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New post 24 Jul 2016, 00:34
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Bunuel,

If the intention is merely to find values of a and b then what is the significance of "Perpendicularity" here? Please explain
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New post 24 Jul 2016, 02:42
Senthil7 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Bunuel,

If the intention is merely to find values of a and b then what is the significance of "Perpendicularity" here? Please explain


The question asks to find the equation of k that is perpendicular to line y=2x and passes through point (a,b). We need to find the values of a and b to answer the question.
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Re: M12-13  [#permalink]

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New post 16 Aug 2016, 13:36
I think that this is a high-quality question and the explanation is very clear. Thanks for posting.
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Re: M12-13  [#permalink]

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New post 17 Aug 2016, 06:30
S1) take different values for a,b that meet the given statement (a=-b) and form lines equations of lines with slope -1/2. You can see that for different values of (a,b), you get different equations.

S2) same as statement 1

Both) Both equations give a point a=1/2 and b=-1/2. Which hivea only one equation 2x+4y=-1

So the ans is C. Hope fully my logic ia correct.

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Re: M12-13  [#permalink]

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New post 07 Oct 2016, 13:30
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)


Check the diagram below:

Image

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png



i want to apologize, as i am very very weak at coordinate geometry.

First, please give a suggestion how can i learn coordinate geometry.

2nd, isn't it obvious in option A that b=5, a= -5; b= 1 a= -1; b=0, a=0; b=1, a= -1?

sorry again if i ask something foolish
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Re: M12-13  [#permalink]

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New post 08 Oct 2016, 03:16
nahid78 wrote:
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)


Check the diagram below:

Image

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png



i want to apologize, as i am very very weak at coordinate geometry.

First, please give a suggestion how can i learn coordinate geometry.

2nd, isn't it obvious in option A that b=5, a= -5; b= 1 a= -1; b=0, a=0; b=1, a= -1?

sorry again if i ask something foolish


For coordinate geometry check this: math-coordinate-geometry-87652.html

For more on this question check here: what-is-the-equation-of-the-line-k-that-is-perpendicular-to-line-y-69481.html
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Re: M12-13  [#permalink]

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New post 20 Dec 2016, 06:23
Hi Bunuel,

A little clarification
equation of line perpendicular to y-2x=0 will be -2y-x+c=0 ??
In order to find equation of a line perpendicular to ax+by+c=0.....interchange coefficients of x and y and change the sign in between....so equation of line perpendicular to ax+by+c=0 will be bx-ay+k=0.
Is this approach correct?
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New post 27 Dec 2016, 11:18
I think this is a high-quality question and I agree with explanation.
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M12-13  [#permalink]

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New post 14 Jan 2017, 10:26
Hi Bunuel,

If the question were to be a slightly different one, would the below Statement be sufficient?

Question stem: Find equation of the line K, perpendicular to a line y= -1/2x and passes through point (a,b).

statement 1) 2a=b

I understood the point that given a slope + either the y-intercept or a specific point, we can plot the specific line. Where I am getting confused is whether we can still identify the line by having the slope and a generic equation such as 2a=b.

When I tried plotting several lines with the slope = 2 and with different y-intercepts, I could not come up with any point that satisfy 2a=b and any other lines except y=2x+0.

Would you please help me clear this point? Thanks!
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Re: M12-13  [#permalink]

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New post 17 Apr 2017, 02:15
I do not understand why we cannot say that stat I implies a=b=0, hence being sufficient while statement II implies basically all consecutive integers, not being sufficient. Can someone please explain?
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New post 17 Apr 2017, 02:20
Kontaxis wrote:
I do not understand why we cannot say that stat I implies a=b=0, hence being sufficient while statement II implies basically all consecutive integers, not being sufficient. Can someone please explain?


a = -b does not necessarily means that a = b= 0. Why not a= 1 and b = -1?
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Re: M12-13  [#permalink]

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New post 10 Sep 2017, 22:59
Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?
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M12-13  [#permalink]

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New post 10 Sep 2017, 23:36
RekhaKulkarni wrote:
Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?


From a=-b and a-b=1, we can get that a = 1/2 and b = -1/2. So, we know that k passes through the point (1/2, -1/2) and is perpendicular to line y = 2x (slope = 2)

The two lines are perpendicular if and only if the product of their slopes is -1, so m*2 = -1 and m = -1/2 (the slope of line k).

Finally, the equation of a straight line that passes through a point \(P_1(x_1, y_1)\) with a slope m is: \(y-y_1=m(x-x_1)\). Substitute: \(y-(-\frac{1}{2})=-\frac{1}{2}(x-\frac{1}{2})\) --> \(y = -\frac{x}{2} - \frac{1}{4}\).

Check below:

Image


For Coordinate Geometry check:

24. Coordinate Geometry



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Re M12-13  [#permalink]

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New post 13 Sep 2017, 23:20
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i read the whole thread, and still didn't understand the explanation. I do understand that we don't need to find the value of a and b, question is why would we discard 1) and 2) as insufficient, what's the reasoning behind it.
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Re M12-13 &nbs [#permalink] 13 Sep 2017, 23:20

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